Dict has key from list

Question

How can I determine if any of the list elements are a key to a dict? The straight forward way is,

for i in myList:
   if i in myDict:
      return True
return False

but is there a faster / more concise way?

Answer 1

#!python
any(x in MyDict for x in MyList)
set(MyList)&set(MyDict)

Answer 2

Assuming you are talking about python, an alternate method of doing this would be:

return len([x for x in myList if x in myDict]) > 0

Answer 3

In addition to any(item in my_dict for item in my_list) from @Ronny's answer:

any(map(my_dict.__contains__, my_list)) # Python 3.x

Or:

from itertools import imap
any(imap(my_dict.__contains__, my_list)) # Python 2.x

Measure relative performance

Cases to consider:

1. Item from the start of list is in in dictionary.
2. Item from the end of list is in dictionary.
3. No items from the list are in dictionary.

Functions to compare (see main.py):

def mgag_loop(myDict, myList):
    for i in myList:
        if i in myDict:
            return True
    return False

def ronny_any(myDict, myList):
    return any(x in myDict for x in myList)

def ronny_set(myDict, myList):
    return set(myDict) & set(myList)

def pablo_len(myDict, myList):
    return len([x for x in myList if x in myDict]) > 0

def jfs_map(my_dict, my_list):
    return any(map(my_dict.__contains__, my_list))

def jfs_imap(my_dict, my_list):
    return any(imap(my_dict.__contains__, my_list))

Results: mgag_loop() is the fastest in all cases.

1. Item from the start of list is in in dictionary.

def args_key_at_start(n):
    'Make args for comparison functions "key at start" case.'
    d, lst = args_no_key(n)
    lst.insert(0, n//2)
    assert (n//2) in d and lst[0] == (n//2)
    return (d, lst)

2. Item from the end of list is in dictionary.

def args_key_at_end(n):
    'Make args for comparison functions "key at end" case.'
    d, lst = args_no_key(n)
    lst.append(n//2)
    assert (n//2) in d and lst[-1] == (n//2)
    return (d, lst)

3. No items from the list are in dictionary.

def args_no_key(n):
    'Make args for comparison functions "no key" case.'
    d = dict.fromkeys(xrange(n))
    lst = range(n, 2*n+1)
    assert not any(x in d for x in lst)
    return (d, lst)

How to reproduce

Download main.py, make-figures.py, run python main.py (numpy, matplotlib should be installed to create plots).

To change maximum size of input list, number of points to plot supply --maxn, --npoints correspondingly. Example:

$ python main.py --maxn 65536 --npoints 16

Answer 4

This was a popular answer to a related question:

>>> if all (k in foo for k in ("foo","bar")):
...     print "They're there!"
...
They're there!

You can adapt it to check if any appears in the dictionary:

>>> if any(k in myDict for k in ("foo","bar")):
...     print "Found one!"
...
Found one!

You can check against a list of keys:

>>> if any(k in myDict for k in myList):
...     print "Found one!"
...
Found one!

Answer 5

Thank you all so much. I tested the performance of all the answers and the fastest was

return len([x for x in myList if x in myDict]) > 0

but I didn't try the "set" answer because I didn't see how to turn it into one line.